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yulyashka [42]
3 years ago
10

Square ABCD has coordinates A(8,7), B(10, 2), and D (3,5). Find the coordinates of C.

Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
7 0

Answer:

Step-by-step explanation:

Area = 1/2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Now by putting all the values in the formula, we will get

Area of triangle = 1/2 [2(0 – (-4)) + (-1)((-4) – (3)) + 2(3 – 0)]

                          = 1/2 [8 + 7 + 6]

                          = 21/2

So, the area of triangle is 21/2 square units.

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If p q r is prime numbers such that pq+r=73, what is the least possible value of p+q+r
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The answer to this question would be: p+q+r = 2 + 17 + 39= 58

In this question, p q r is a prime number. Most of the prime number is an odd number. If p q r all odd number, it wouldn't be possible to get 73 since
odd x odd + odd= odd + odd = even
Since 73 is an odd number, it is clear that one of the p q r needs to be an even number. 

There is only one odd prime number which is 2. If you put 2 in the r the result would be:
pq+2= 73
pq= 71
There will be no solution for pq since 71 is prime number. That mean 2 must be either p or q. Let say that 2 is p, then the equation would be: 2q + r= 73

The least possible value of p+q+r would be achieved by founding the highest q since its coefficient is 2 times r. Maximum q would be 73/2= 36.5 so you can try backward from that. Since q= 31, q=29, q=23 and q=19 wouldn't result in a prime number r, the least result would be q=17
r= 73-2q
r= 73- 2(17)
r= 73-34=39

p+q+r = 2 + 17 + 39= 58
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